Theorem 4.5.99D6B5CF5E0CF

Proof. For $C\Subset W\Subset \Omega $ containing $a$, Harnack comparison in the pole variable, or sweeping onto a compact neighborhood of $C$, bounds $G(y,x)/G(y,a)$ uniformly for $x\in C$, $y\notin W$. For the sweeping proof, both pole functions equal their compact reductions off that neighborhood: they are bounded harmonic there and have the same boundary values outside a polar exception, so compactified uniqueness applies. Their boundary values on its compact interior boundary have uniform positive lower and finite upper bounds, giving the ratio bound. A general compact $C$ is enlarged to include $a$. Since $b=G(\cdot ,a)$ off a compact set, these bounds also hold for $k$. Normal convergence and the derivative estimates for bounded harmonic families show that every boundary kernel is harmonic, and that convergence in the compactification is locally uniform in $x$. This proves joint continuity at boundary points. Interior joint continuity is that of $G$, and $K_ y(a)=1$ on the boundary follows from the same identity off the cutoff compact. Positivity follows from the harmonic minimum principle.

Interior kernels are distinguished by their poles. A boundary kernel has no interior pole. For two boundary points, proportionality and the value one at $a$ imply equality of every defining coordinate, hence equality of the points. More explicitly, $b=G\lambda _0$ with compactly supported $\lambda _0$; therefore $\int K(y,x)d\lambda _0(x)=1$, first for interior $y$, then on the boundary by joint continuity near that compact support. This supplies the normalization and the ratio representation of defining coordinates used in Helms 12.4 as well. The countable coordinates in the displayed formula separate points: density and extended continuity in $x$ recover every kernel. They define a continuous injection of compact $M$ into a countable cube, hence a homeomorphism onto its image. The uniformly convergent sum metric on that cube induces the given topology. No sequential compactness is assumed before this argument. $\square$